# Prove the distributive law $a(b+c)=ab+ac$ for real numbers?

I have always taken these kinds of things for granted. Well of course $a(b+c)=ab+ac$! But why? The thought randomly popped in my head, and I realized that I could not prove it.

Perhaps we should take some time to prove these "obvious" things. I feel like if I cannot prove it, then I do not fully understand it. I would really like a proof of this most basic law, because I have been pondering over it for quite a while. Thanks!

Edit: This proof is for real numbers.

• For what kind of objects $a$, $b$, $c$ do you want to prove this? – k.stm May 8 '14 at 6:48
• There are two common approaches to defining the real numbers I know of. One of them is as a totally-ordered, Dedekind-complete field, and distributivity is one of the field axioms, so you just get that distributivity holds by definition. (Proving the real numbers exist with this definition is then a more interesting task.) The other approach constructs the reals from the rationals, in which case distributivity can be proven using the techniques of analysis and the distributivity of the rationals. The distributivity of the rationals can then be proven from the distributivity of the integers, – user2357112 May 8 '14 at 7:00
• which can then be proven from the definition of the integers. – user2357112 May 8 '14 at 7:01
• I always have a problem when trying to answer this kind of question. (─‿‿─) – Anastasiya-Romanova 秀 May 8 '14 at 17:29
• For an intuition using geometry, I'd found this video useful. – Anant May 11 '14 at 17:52

Henry Reich - creator of minutephysics - gave a geometrical intuition for the distributive law in this video. I've pasted a snapshot from that below.

For a formal proof, I defer to Christian Blatter's answer :)

(This is to @JChau who "simply want[s] a full proof from someone why the property $a(b+c)=ab+ac$ is true for real numbers".)

A full proof involves a construction of the reals from the Peano axioms. Landau's classic Grundlagen der Analysis from 1930 takes 90 pages to do this. If you are willing to accept the distributive law for integers you can bring this down to a dozen pages or so. There are many constructions of the reals around; and all of them require about the same amount of work. A comparatively short version can be found in chapter 1 of Baby Rudin.

For natural numbers, the "proof" would look like this.

Since $m\cdot n$ is just $n+n+\dots+n$, repeated $m$ times, this means that $$(p+q)\cdot n = n+n+\cdots+n,$$ repeated $p+q$ times, but it is also equal to $$\underbrace{(n+n+\cdots n)}_{p}+\underbrace{(n+n+\cdots +n)}_{q}$$ which is $p\cdot n + q\cdot n$.

For rational numbers, the expression follows from the definition $$\frac pq\cdot \frac rs = \frac{pr}{qs}.$$

For the real numbers, you have to follow the construction of real numbers from rational ones using Dedekind cuts, for example.

• "You can't really prove te law for reals" - of course you can! Why not? – user2357112 May 8 '14 at 6:59
• @user2357112 Yes, poorly expressed. I fixed my answer. – 5xum May 8 '14 at 7:00